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Three-compartment migration conserves weighted mass

Example

Consider a closed three-vertex path with orientation 1→2→3, masses (1,2,3), conductances (1,2), homogeneous diffusion, and initial state u(0)=(2,1,4)T. Its conserved weighted mass is 16. The constant state (8/3,8/3,8/3)T is an equilibrium with that same weighted mass, and the exact trajectory from the stated nonnegative initial state stays entrywise nonnegative. No convergence-to-equilibrium claim is made.

The diagram records the orientation, conductances, and vertex masses; edge lengths are schematic and are not used in the matrix calculation.

\begin{center} \begin{tikzpicture}[>=stealth, every node/.style={font=\small}] \node[circle,draw] (v1) at (0,0) {1}; \node[circle,draw] (v2) at (2.1,0) {2}; \node[circle,draw] (v3) at (4.2,0) {3}; \draw[->] (v1) -- node[above] {c1=1} (v2); \draw[->] (v2) -- node[above] {c2=2} (v3); \node[below=4pt] at (v1) {m1=1}; \node[below=4pt] at (v2) {m2=2}; \node[below=4pt] at (v3) {m3=3}; \end{tikzpicture} \end{center}

Facts & Assumptions

Given: The oriented path, M=diag⁡(1,2,3), C=diag⁡(1,2), D=∅, and the stated initial vector.

[F1]

The closed network equation is Mu′+LCu=0, with LC=BCBT and weighted mass 1TMu (Finite weighted network diffusion data).

[F2]

For a closed network, 1TMu is constant under homogeneous diffusion; constants on a positive-conductance connected component lie in ker⁡LC (Energy, kernel, and weighted mass balance for a finite network).

[F3]

The homogeneous exact semigroup is defined by the finite matrix exponential and preserves nonnegative data (Positive graph-diffusion semigroup and Euler maps).

Verification

Given: The three-compartment matrices, initial state, and homogeneous network equation in the Example.

1.1F1givenalgebra

With the stated orientation, B=(−101−101),LC=BCBT=(1−10−13−20−22). The graph is connected and both conductances are positive.

1.2F1F2givenalgebra

The initial weighted mass is 1TMu(0)=2+2+12=16. By [F2] it is constant for all times for which the homogeneous solution is evaluated.

2.1F1F2step 1.1algebra

The vector 1=(1,1,1)T has LC1=0 by the displayed matrix in step 1.1, and its weighted mass at value 8/3 is (8/3)(1+2+3)=16. Thus the stated constant state is an equilibrium with the same weighted mass.

3.1F3givenstep 1.1∎

Since u(0)≥0, [F3] implies the exact semigroup trajectory remains entrywise nonnegative for every t≥0.

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